A) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]
B) \[{{b}^{3}}+{{a}^{2}}c+a{{c}^{2}}=3abc\]
C) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2abc\]
D) \[{{b}^{2}}c+{{c}^{2}}a+{{a}^{2}}b=abc\]
Correct Answer: B
Solution :
Let roots are \[\alpha \]and \[{{\alpha }^{2}}\]. \[\therefore \,\,\,\,\,\,\alpha +{{\alpha }^{2}}=-\frac{b}{a}\]and \[{{\alpha }^{3}}=\frac{c}{a}\] \[\Rightarrow \] \[(\alpha +{{\alpha }^{2}})=-{{\left( \frac{b}{a} \right)}^{3}}\] \[\Rightarrow \] \[{{\alpha }^{3}}+{{\alpha }^{6}}+3{{\alpha }^{3}}(\alpha +{{\alpha }^{2}})=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \] \[\,\frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}+\frac{3c}{a}\left( -\frac{b}{a} \right)=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \] \[\,\frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}-\frac{3bc}{{{a}^{2}}}=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\therefore \,\,\,\,{{b}^{3}}+{{a}^{2}}c+a{{c}^{2}}-3abc=0\]
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